poj 3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25370 Accepted: 11418

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 61 42 63 122 7

Sample Output

23

题目链接：http://poj.org/problem?id=3624

题目大意：有n只魔力手镯，已知每只手镯的重量和魔力值，求总重量不超过m的所有手镯的最大魔力总值。

解题思路：经典背包问题的简单题。dp[ v ]表示总重量在 v 以内的所有手镯的最大魔力总值。

方程为：  dp[ v ] = dp [ v -w[ i ] ] + d [ i ]  写的时候用了结构，但用两个数组记录重量和价值就可以了。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int dp[12890];//dp[i]表示在重量不超过i的情况下的最大价值struct A{    int w,d;//重量和价值}a[3410];int main(){    int i,j,n,m,ans=0;memset(dp,0,sizeof(dp));    scanf("%d%d",&n,&m);    for(i=0;i<n;i++)        scanf("%d%d",&a[i].w,&a[i].d);for(i=0;i<n;i++)      //i从前往后，重量从大到小根据a[i]更新重量j以内的最大价值量for(j=m;j>=a[i].w;j--)      // 第一次遍历，重量不小于a[0].w的dp值全更新为a[0].d的值，依次遍历不断更新。if(dp[j]<dp[j-a[i].w]+a[i].d)     dp[j]=dp[j-a[i].w]+a[i].d;printf("%d\n",dp[m]);    return 0;}